Question

$\boxed{\bigstar\boxed{\mathbb{MATH~~TEST}}\bigstar} \\\\ \mathrm{If~x=4~and~y=27,then~\boxed{\mathrm{ \frac{7x^2y^{ \frac{5}{6} } }{(x^{ \frac{5}{4}}-6y^{- \frac{1}{3} }) \sqrt{x} ^3} }}}=... \\\\ \mathrm{A.1(1+2 \sqrt{2})18 \sqrt{3} } \\\ \mathrm{B.(1+2 \sqrt{2})27 \sqrt{2} } \\\ \mathrm{C.(1+2 \sqrt{2} )9 \sqrt{2} } \\\ \mathrm{D.(1+2 \sqrt{2}) 9 \sqrt{3} } \\\ \mathrm{E.(1+2 \sqrt{2} )27 \sqrt{3} }$

Answer 1

D.
==========
x=4            √x =2                 x^(5/4) = 2^(5/2) = √32
y = 27       y^(-1/3) = 1/y^(1/3) = 1/3       y^(5/6) = 3^(5/2) = √243

$\frac{7x^2 y^\frac{5}{6}}{(x^\frac{5}{4}-6y^{-\frac{1}{3}}) (\sqrt{x})^3}\\\\=\frac{7*4^2*\sqrt{243}}{(\sqrt{32}-6*\frac{1}{3})2^3}\\\\=\frac{7*16*9\sqrt{3}}{(4\sqrt2-2)8}\\\\=\frac{63\sqrt3}{2\sqrt2-1},\ \ multiply\ by \ <span>(2\sqrt2+1)\ in\ Nr\ and\ Dr </span>\\\\=\frac{63\sqrt3*(2\sqrt2+1)}{8-1}\\\\=9\sqrt3\ (2\sqrt2+1)$